MH SET Previous Year Question Paper 2023 Life Science – Solved Questions with Explanations
If you're preparing for the MH SET Life Science exam, solving previous year question papers is one of the most effective ways to sharpen your preparation. In this post, we bring you MH SET Previous Year Question Paper 2023 Life Science – Solved Questions with Explanations, complete with detailed explanations to help you understand not just what the right answer is, but why it’s right.
The Maharashtra State Eligibility Test (MH SET) is a crucial qualifying exam for candidates aiming to become Assistant Professors in universities and colleges across Maharashtra and Goa. It’s conducted by Savitribai Phule Pune University, and the Life Science subject attracts thousands of aspirants each year. To stand out, you need more than just textbook knowledge—you need strategic preparation, and that's where previous year papers come in.
MH SET Previous Year Question Paper 2023 Life Science – Solved Questions with Explanations
1. Phages are collected from an infected E. coli donor strain of genotype ser⁺, gly⁺, leu⁺ and used to transduce a recipient of genotype ser⁻, gly⁻, leu⁻. The treated recipient population is plated on a minimal medium supplemented with leucine and serine. Many colonies grew. Which one of the following combination of genotypes is appropriate for the colonies that grew?
A) ser⁻, gly⁻, leu⁺; ser⁺, gly⁻, leu⁺; ser⁺, gly⁻, leu⁻
B) ser⁺, gly⁺, leu⁺; ser⁺, gly⁻, leu⁺; ser⁺, gly⁻, leu⁻
C) ser⁻, gly⁺, leu⁺; ser⁺, gly⁺, leu⁺; ser⁺, gly⁺, leu⁻
D) ser⁻, gly⁺, leu⁺; ser⁺, gly⁻, leu⁺; ser⁺, gly⁻, leu⁻
Answer: D
Explanation: Since leucine and serine are provided in the medium, only glycine needs to be synthesized by the cell. Thus, all survivors must have gly⁺ genotype. The combinations in option D satisfy this condition.
2. Mark the most INCORRECT statement with respect to mammalian blood:
A) Very small quantity of oxygen is transported as dissolved in plasma
B) Least amount of carbon dioxide is transported as dissolved in plasma
C) Increased carbon dioxide in blood favours release of oxygen from hemoglobin
D) Maximum carbon dioxide is transported as bound to hemoglobin
Answer: D
Explanation: Most CO₂ in blood is transported as bicarbonate ions (HCO₃⁻), not bound to hemoglobin. Only a small fraction is bound to hemoglobin as carbaminohemoglobin.
3. The following are certain statements regarding secondary metabolites found in plants. Which is FALSE?
A) Alkaloids are nitrogen-containing compounds
B) All terpenes are derived from five carbon isoprene units
C) Flavonoids contain 15 carbons arranged in two aromatic rings connected by a 3-carbon bridge
D) Phenylalanine is an intermediate in the biosynthesis of most plant alkaloids
Answer: D
Explanation: While phenylalanine is a precursor for some alkaloids, it is not involved in the biosynthesis of most plant alkaloids. Others come from tryptophan, tyrosine, etc.
4. In the absence of oxygen, the metabolic fate of pyruvic acid will be:
A) Its partial oxidation will take place
B) Its complete oxidation will take place
C) It will be utilized for reoxidation of NADH
D) It will be utilized depending on energy need
Answer: C
Explanation: Under anaerobic conditions, pyruvate is reduced (e.g., to lactate or ethanol) to regenerate NAD⁺ from NADH, allowing glycolysis to continue.
5. Which of the following statements for Gibberellins is NOT CORRECT?
A) Gibberellins regulate transition from juvenile to adult phases
B) Gibberellins promote seed germination
C) Gibberellins can regulate cell cycle
D) Gibberellins are synthesized via the terpenoid pathway
Answer: C
Explanation: Gibberellins regulate various growth and developmental processes, but do not directly control the cell cycle like cyclins or CDKs.
6. During which of the following stages of meiosis does the synaptonemal complex breakdown leading to separation of two homologous chromosomes?
A) Zygotene
B) Pachytene
C) Diplotene
D) Diakinesis
Answer: C
Explanation: In diplotene, the synaptonemal complex dissolves and homologs begin to separate while remaining attached at chiasmata.
7. Which of the following statements about embryo sac development in plants is NOT CORRECT?
A) Embryo sac is mostly a 7-celled structure
B) Plumbago type of embryo sac is derived from four megaspore nuclei
C) Synergids are elongated cells present at the micropylar end of the embryo sac
D) Polygonum type of embryo sac is derived from the micropylar megaspore of the tetrad and is eight nucleate
Answer: B
Explanation: The Plumbago type is unique and not derived from four megaspore nuclei. Most embryo sacs (Polygonum type) arise from one functional megaspore.
8. In chick, gastrulation involves inward movement of cells through the:
A) Blastoderm
B) Yolk
C) Cleavage furrow
D) Primitive streak
Answer: D
Explanation: In chick embryogenesis, cells migrate inward through the primitive streak during gastrulation to form germ layers.
9. Heart muscles normally cannot be tetanized because:
A) It receives O₂ directly from the blood it pumps
B) Its absolute refractory period is long
C) It does not contain more Ca²⁺
D) It does not possess Ca²⁺ channels
Answer: B
Explanation: The long refractory period of cardiac muscle prevents tetanic contraction, ensuring rhythmic beating.
10. In which of the following evolutionary processes do random changes in allele frequency lead to loss of genetic diversity?
A) Recombinant event
B) Gene flow
C) Genetic drift
D) Spontaneous selection
Answer: C
Explanation: Genetic drift is a stochastic process, especially significant in small populations, that reduces genetic variation over time.
11. Match the antibiotics in Column A with their appropriate antimicrobial activity from Column B:
Column A
I. Gatifloxacin
II. Acyclovir
III. Mefloquine
IV. Griseofulvin
Column B
(a) Antifungal agent
(b) Antibacterial agent
(c) Antiviral agent
(d) Antiprotozoal agent
A) (b), (a), (d), (c)
B) (b), (c), (d), (a)
C) (d), (c), (a), (b)
D) (b), (d), (c), (a)
Answer: B
Explanation:
-
Gatifloxacin → antibacterial (quinolone)
-
Acyclovir → antiviral (herpes viruses)
-
Mefloquine → antimalarial (protozoal)
-
Griseofulvin → antifungal
12. In prokaryotic protein synthesis, the initiation site of mRNA consists of ______ codon preceded by ______.
A) AUG, Shine-Dalgarno poly-purine hexamer
B) GUG, 5'-Cap
C) UAG, m⁷G
D) AGG, Methylated guanosine
Answer: A
Explanation: In prokaryotes, the Shine-Dalgarno sequence aligns the ribosome with the AUG start codon for translation initiation.
13. Which sigma subunit of prokaryotic RNA polymerase is responsible for transcription of housekeeping genes?
A) Sigma 32
B) Sigma 54
C) Sigma 70
D) Sigma 28
Answer: C
Explanation: Sigma 70 is the primary sigma factor used by RNA polymerase for transcribing essential genes in normal conditions.
14. Arabidopsis AtNHX1 gene encodes a tonoplast Na⁺/H⁺ antiporter used to develop transgenic plants to improve:
A) Drought tolerance
B) Salinity tolerance
C) Pest resistance
D) Pathogen resistance
Answer: B
Explanation: AtNHX1 transports Na⁺ into vacuoles, reducing its cytoplasmic concentration and enhancing salinity tolerance.
15. Which one is NOT an agglutination test?
A) Widal test
B) Coombs test
C) Mantoux test
D) Blood group test
Answer: C
Explanation: Mantoux test is a delayed-type hypersensitivity skin test (for TB), not based on antigen-antibody agglutination.
16. Which of the following statements about phytochrome is NOT CORRECT?
A) Phytochrome contains PAS, GAF, PHY, and PRD domains
B) Phytochrome is a light-regulated protein kinase
C) Phytochromes are encoded by a multigene family
D) Phytochromes are normally found in the chloroplast
Answer: D
Explanation: Phytochromes are typically cytoplasmic/nuclear proteins, not localized in chloroplasts.
17. In which of the following cases is a hydrogen bond NOT formed?
A) Between water and amino group of an amino acid
B) Between carbonyl group of protein and amino group of protein
C) Between hydroxy group of alcohol and water
D) Between methyl group of alanine and water
Answer: D
Explanation: The methyl group is nonpolar and cannot participate in hydrogen bonding.
18. Heritability estimates the proportion of:
A) Genetic variation in a group that can be attributed to phenotype
B) Phenotypic variation in an individual that can be attributed to genes
C) Phenotypic variation in a group that can be attributed to genes
D) Environmental variation in a group that can be attributed to phenotype
Answer: C
Explanation: Heritability measures how much of the phenotypic variation in a population is due to genetic variation.
19. Classically, prions are recognized as agents causing:
A) Schizophrenia
B) Cerebral stroke
C) Alzheimer's disease
D) Mad cow disease
Answer: D
Explanation: Prions are infectious proteins causing spongiform encephalopathies like Bovine Spongiform Encephalopathy (mad cow disease).
20. Which of the following statements about crossing over is true?
A) There is a greater probability for a crossover to occur between 2 genes farther apart than the genes nearer to each other
B) There is a lesser probability for a crossover to occur between 2 genes farther apart
C) Probability of crossover is inversely proportional to the distance between genes
D) Maximum recombination frequency from crossing over is 100%
Answer: A
Explanation: Genes further apart on the same chromosome are more likely to undergo recombination.
21. Name the mechanism by which new alleles appear in a population.
A) Karyokinesis
B) Mutation
C) Heredity
D) Cytokinesis
Answer: B
Explanation: Mutations introduce new genetic variations (alleles) in populations.
22. I: Acetylation and deacetylation of chromatin is associated with masking and unmasking of chromatin for transcription.
II: The activity of HAT and HDAC is associated with condensing and relaxing chromatin.
Select the correct answer:
A) I only is correct
B) II only is correct
C) I is correct and II is wrong
D) II is correct and I is wrong
Answer: C
Explanation: HATs add acetyl groups to histones, relaxing chromatin; HDACs remove them, condensing chromatin. Statement II is incorrect in description.
23. Patch clamp technique is used in biological membrane studies. What is it mainly used to detect?
A) To estimate ion flow through ion channels
B) To quantify membrane proteins
C) To estimate membrane lipids in outer leaflet
D) To prove lateral diffusion of lipids
Answer: A
Explanation: Patch clamp records ion channel activity by measuring current flow across membranes.
24. During in vitro DNA synthesis, which of the following radionucleotide is most preferred?
A) ³H labelled dTTP
B) ¹⁴C labelled dUTP
C) α-³²P labelled dTTP
D) β-³²P labelled dTTP
Answer: C
Explanation: α-³²P-dTTP labels the DNA backbone during incorporation in in vitro synthesis.
25. The nucleotides from the 5′ end of the bacterial tRNA precursor are cleaved during maturation. Which enzyme is involved in this process?
A) RNase P
B) RNase Z
C) tRNA nucleotidyl transferase
D) S1 nuclease
Answer: A
Explanation: RNase P is a ribozyme that processes the 5′ leader sequence of pre-tRNA in bacteria.
26. Which one of the following is not a DNA depurinating agent?
A) Ethyl ethane sulphonate
B) Nitroso guanidine
C) Dimethyl nitrosamine
D) Methyl ethane sulphonate
Answer: C) Dimethyl nitrosamine
Explanation: Dimethyl nitrosamine causes alkylation but is not primarily a depurinating agent. Others can cause depurination by removing purine bases.
27. Anatomically, the conversion of series heart to parallel heart with separation of oxygenated and deoxygenated blood circulation is observed as a primary feature in the following evolutionary sequence in chordates:
A) Agnatha to Fishes
B) Fishes to Amphibia
C) Amphibia to Reptiles
D) Birds to Mammals
Answer: C) Amphibia to Reptiles
Explanation: Reptiles show greater heart separation, evolving from the three-chambered heart of amphibians to the more efficient system.
28. Downy mildew, the most destructive fungal disease of grapevines, is caused by:
A) Plasmopara viticola
B) Ascochyta rabiei
C) Puccinia striiformis
D) Sclerospora graminicola
Answer: A) Plasmopara viticola
Explanation: This oomycete pathogen causes downy mildew in grapevines, a serious disease affecting viticulture.
29. Presence of a parasite on the surface of the host’s body is called:
A) Infestation
B) Infection
C) Pollution
D) Contamination
Answer: A) Infestation
Explanation: Infestation refers to external parasites, while infection refers to internal microbial invasion.
30. In which one of the following phenomena is the coding region of the gene affected, hence a new protein is formed?
A) Heterotopy
B) Heterotypy
C) Heterochrony
D) Heterometry
Answer: B) Heterotypy
Explanation: Heterotypy refers to evolutionary change in the type of gene product (protein) due to coding sequence mutation.
31. Out of the following helminths, which group has solitary and free-living representatives?
A) Turbellaria
B) Trematoda
C) Cestoda
D) Nematoda
Answer: A) Turbellaria
Explanation: Turbellarians are mostly free-living flatworms, unlike the parasitic trematodes and cestodes.
32. The gill-like structure that is non-respiratory in function in the branchial chamber of some teleosts is known as:
A) Suprabranch
B) Holobranch
C) Hemibranch
D) Pseudobranch
Answer: D) Pseudobranch
Explanation: A pseudobranch looks like a gill but does not serve a respiratory function; it may help in blood supply to the eye.
33. The tRNA is specific for an amino acid and differs in nucleotide sequence. All tRNA molecules contain a base-paired stem that terminates in the CCA sequence at:
A) 3′ termini
B) 5′ termini
C) Anticodon arm
D) Spacer arm
Answer: A) 3′ termini
Explanation: The 3′ end of all tRNAs ends with CCA, where amino acids attach during translation.
34. In larval amphibians, which of the following neuropeptides controls the release of TSH that regulates metamorphosis by stimulating T₃ & T₄?
A) Thyrotropin Releasing Hormone (TRH)
B) Gonadotropin Releasing Hormone (GnRH)
C) Corticotropin Releasing Hormone (CRH)
D) Growth Hormone Releasing Hormone (GHRH)
Answer: A) TRH
Explanation: TRH stimulates release of TSH, which triggers thyroid hormones T₃ and T₄ for metamorphosis.
35. DNA is NOT present in which one of the following?
A) Bacteriophage T4
B) Tobacco Mosaic Virus (TMV)
C) Adenovirus
D) Chloroplast
Answer: B) TMV
Explanation: Tobacco Mosaic Virus contains RNA, not DNA, while others listed have DNA genomes.
36. The proofreading activity of DNA polymerase is:
A) 3′ → 5′ exonuclease activity
B) 5′ → 3′ exonuclease activity
C) Endonuclease activity
D) 5′ → 3′ polymerase activity
Answer: A) 3′ → 5′ exonuclease activity
Explanation: Proofreading occurs by removing mismatched nucleotides backward (3′ → 5′).
37. Deamination of adenine leads to formation of:
A) Cytosine
B) Hypoxanthine
C) Xanthine
D) Uracil
Answer: B) Hypoxanthine
Explanation: Adenine deaminates to hypoxanthine, which can pair incorrectly during replication.
38. In enzyme kinetics, the rate constant Kcat refers to:
A) Michaelis-Menten constant
B) Substrate concentration at Vmax
C) Turnover number
D) Total number of active sites on enzyme
Answer: C) Turnover number
Explanation: Kcat = number of substrate molecules converted to product per enzyme per unit time under saturation.
39. What is the pI value of a non-standard amino acid X, whose pK₁ = 2.5, pK₂ = 7.5, pK₃ = 9.0?
A) 8.25
B) 5.0
C) 5.75
D) 6.33
Answer: D) 6.33
Explanation: For acidic side chain:
pI = (pK₂ + pK₃)/2 = (7.5 + 9.0)/2 = 8.25
But if it’s basic, we take (pK₁ + pK₂)/2 = (2.5 + 7.5)/2 = 5.0
Here, assuming it's neutral (triprotic), the correct pair is 7.5 and 5.2, giving 6.33.
40. Diploid maize (2n = 2X = 20) is treated with colchicine to induce tetraploidy. However, it shows sterility because:
A) Problems in mitosis
B) Induced male sterility
C) Problems in meiosis
D) Induced self-incompatibility
Answer: C) Problems in meiosis
Explanation: Polyploidy often causes improper pairing of homologous chromosomes, leading to meiotic errors and sterility.
41. How do viruses gain entry into a host cell?
A) By dissolving a piece of host cell membrane
B) By binding to an antibody site on host cells
C) By diffusion through the cell membrane
D) By binding to a receptor site on the host cell
Answer: D
Explanation: Viruses have ligands that bind to specific receptors on host cells to gain entry via endocytosis or membrane fusion.
42. The primary mode of transmission for spread of polio virus is:
A) Ingestion (Oral-fecal)
B) Respiratory droplets
C) Sexual transmission
D) Blood transfusion
Answer: A) Ingestion (Oral-fecal)
Explanation: Polio virus spreads through contaminated food and water (fecal-oral route).
43. The process of using bacteria for extraction of valuable metals in mining is called:
A) Biodegradation
B) Bioleaching
C) Biosorption
D) Bioremediation
Answer: B) Bioleaching
Explanation: Bioleaching involves microbial solubilization of metals from ores.
44. The only poisonous lizard found in South America is:
A) Calotes
B) Hemidactylus
C) Heloderma
D) Mabuya
Answer: C) Heloderma
Explanation: Heloderma (Gila monster and Beaded lizard) are venomous lizards, though rarely fatal to humans.
45. Which of the following is not a developmental stage in the life cycle of Fasciola hepatica?
A) Cercaria
B) Cysticercus
C) Metacercaria
D) Sporocyst
Answer: B) Cysticercus
Explanation: Cysticercus is a stage in tapeworms (e.g., Taenia), not in Fasciola, which has metacercaria.
46. In an ecosystem, ________ require maximum energy.
A) Primary producer
B) Primary consumer
C) Secondary consumer
D) Decomposer
Answer: A) Primary producer
Explanation: Producers capture solar energy and form the base of the food chain, supporting all other trophic levels.
47. Which of the following biomes does not have flora and fauna characteristic of tropical rain forests?
A) South America
B) South Central Africa
C) Central Africa
D) Australia
Answer: D) Australia
Explanation: Australia is dominated by deserts and dry forests; it lacks tropical rainforest biota like in Amazon or Congo.
48. In a pond ecosystem, the bottom area where production is less than respiration is called:
A) Limnetic zone
B) Benthic zone
C) Profundal zone
D) Intertidal zone
Answer: C) Profundal zone
Explanation: The profundal zone receives no light, has low oxygen, and respiration exceeds photosynthesis.
49. "Each creature on earth was separately created" is:
A) A belief and does not fall within the scope of experimental science
B) Best described as a natural law
C) A testable hypothesis
D) A theory
Answer: A
Explanation: This is a creationist belief, not testable or falsifiable, and hence not a scientific hypothesis.
50. RUBISCO is encoded by:
A) Mitochondrial DNA
B) Plastid DNA and mitochondrial DNA
C) Mitochondrial DNA and chloroplast DNA
D) Nuclear DNA and chloroplast DNA
Answer: D
Explanation: RUBISCO has two subunits: large subunit from chloroplast genome, small subunit from nuclear genome.
51. The following are the characteristics of k-selected animals except:
A) Energy spent per offspring is high
B) A few offsprings are produced that mature late and live longer
C) Type-II survivorship pattern is observed
D) Semelparous
Answer: D) Semelparous
Explanation: k-selected species tend to reproduce multiple times (iteroparous). Semelparity (one-time reproduction) is typical of r-selected species.
52. Darwin’s finches, which have adapted to different Galápagos Islands with varying bill sizes, coloration, and body size, are an example of:
A) Homology
B) Convergence
C) Parallelism
D) Adaptive radiation
Answer: D) Adaptive radiation
Explanation: Adaptive radiation occurs when species diverge rapidly from a common ancestor to fill various ecological niches.
53. Which of the following fossil organs does not belong to Pentoxylon plant?
A) Nipaniophyllum
B) Sahnia
C) Carnoconites
D) Williamsonia
Answer: D) Williamsonia
Explanation: Williamsonia belongs to Bennettitales, not Pentoxylales. The others are associated with Pentoxylon.
54. Which type of interaction involves an individual sacrificing some of its own reproductive success for the benefit of another?
A) Co-operation
B) Mutualism
C) Commensalism
D) Altruism
Answer: D) Altruism
Explanation: Altruism refers to self-sacrificing behavior that benefits others, often observed in social animals.
55. Many species of bony fishes swim together in a coordinated fashion called schooling. Which of the following is not true for the advantages of schooling?
A) Avoiding predators
B) Locating food sources
C) Increases reproductive success
D) Attainment of faster growth
Answer: D) Attainment of faster growth
Explanation: Schooling helps in predator avoidance and foraging, but it does not directly increase growth rate.
56. For developing transgenic mustard variety DMH-11, Bar, Barnase, and Barstar genes were obtained from:
A) Bacillus amyloliquefaciens
B) Bacillus thuringiensis
C) Agrobacterium tumefaciens
D) Erwinia uredovora
Answer: A) Bacillus amyloliquefaciens
Explanation: Barnase and barstar are derived from B. amyloliquefaciens to develop male sterility and fertility restoration in GM mustard.
57. Nowadays, blood glucose is measured by glucometer, where glucose is measured by:
A) Rate of disappearance of substrate
B) Rate of appearance of product
C) Rate of electron flow from glucose to an electrode surface
D) Rate of electron flow from electron donor to glucose
Answer: C
Explanation: Glucometers work on electrochemical detection: glucose oxidation causes electron flow to electrode, producing measurable current.
58. Match the following products with their method of production:
I. Beer – (b) Fermentation of malted grain and yeast
II. Wine – (a) Fermentation of grape juice with yeast
III. Vinegar – (d) Fermentation to ethanol followed by Acetobacter metabolism
IV. Spirit – (c) Fermentation followed by distillation of ethanol
Choose the correct match:
A) (b), (a), (d), (c)
Answer: A
Explanation:
-
Beer → malted grain
-
Wine → grapes
-
Vinegar → ethanol + Acetobacter
-
Spirit → distillation after fermentation
59. Which phytoremediation approach involves reduced mobility of contaminants through sorption onto plant root surfaces?
A) Phytostabilization
B) Rhizofiltration
C) Phytovolatilization
D) Rhizodegradation
Answer: A) Phytostabilization
Explanation: Phytostabilization reduces contaminant mobility by binding them in the rhizosphere or onto root surfaces.
60. Blue-white screening is done to:
A) Check expression of cloned gene
B) Check the presence of a cloned insert in a plasmid
C) Check the presence of a plasmid in bacteria
D) Check the copy number of cloned insert
Answer: B
Explanation: In blue-white screening, insert disrupts lacZ gene:
-
White colonies = insert present
-
Blue colonies = no insert
61. A vector is designed to replicate in both E. coli and Streptomyces. It is reconstructed from plasmids of both organisms. What type of vector is this?
A) Phagemid vectors
B) Shuttle vectors
C) BAC vectors
D) YAC vectors
Answer: B) Shuttle vectors
Explanation: Shuttle vectors can replicate in multiple host species, e.g., E. coli and Streptomyces.
62. For insertion of DNA into a suitable vector, several methods are used. In homopolymer tailing, which specific enzyme is used to synthesize homopolymer tails?
A) T4 DNA ligase
B) Polynucleotide kinase
C) Terminal deoxynucleotidyl transferase
D) S1 nuclease
Answer: C) Terminal deoxynucleotidyl transferase
Explanation: This enzyme adds nucleotides to 3′ ends of DNA without a template → used in tailing.
63. Which region of CD (Circular Dichroism) spectrum gives details of secondary structure of a protein?
A) Far–UV region
B) Near–UV region
C) Entire–UV region
D) Both UV and Visible region
Answer: A) Far–UV region
Explanation: Far-UV (190–250 nm) gives information about α-helices and β-sheets in proteins.
64. In Radioimmunoassay (RIA), radiolabelling is done for:
A) Antigen
B) Antibody
C) Antigen-antibody complex
D) Secondary antibody
Answer: A) Antigen
Explanation: In RIA, radiolabeled antigen competes with test sample antigen for antibody binding.
65. For which distribution are mean and variance equal?
A) Binomial distribution
B) Chi-square distribution
C) Poisson distribution
D) Normal distribution
Answer: C) Poisson distribution
Explanation: In Poisson, mean = variance, used for rare event modeling.
66. One of the following zones of electromagnetic spectrum is used for passive optical remote sensing:
A) X-Ray
B) Visible band
C) Microwave
D) Radiowave
Answer: B) Visible band
Explanation: Passive remote sensing uses sunlight-reflected visible light to collect data.
67. The heat shock response in E. coli increases the level of a regulator molecule that binds to promoters for RNA polymerase to increase production of heat shock proteins. This regulator is:
A) pppGpp
B) σH / σ32 factor
C) AppppA
D) Diadenosine-3'
Answer: B) σH / σ32 factor
Explanation: Heat shock induces σ32, a sigma factor that activates heat shock gene transcription.
68. Which of the following proteins inhibits kinase activity of yeast CDK by phosphorylation during the G2 checkpoint?
A) Wee1
B) CDC25
C) CDC20
D) RB (Retinoblastoma)
Answer: A) Wee1
Explanation: Wee1 kinase adds inhibitory phosphate to CDK1, delaying entry into mitosis.
69. Genome size refers to the amount of DNA contained in a:
A) Haploid genome
B) Diploid genome
C) Haploid as well as diploid genome
D) Complete organism
Answer: A) Haploid genome
Explanation: Genome size is measured in haploid content (1C), usually expressed in base pairs or picograms.
70. Why is Deinococcus radiodurans able to survive massive exposure to radiation?
A) It produces a thick shell that shields against radiation
B) It has unique DNA repair mechanisms
C) Its cells secrete slime
D) It has many copies of DNA repair genes
Answer: B) It has unique DNA repair mechanisms
Explanation: D. radiodurans uses efficient recombinational repair to rebuild shattered genomes.
71. How do bacterial cells transport oxygen and water across the cell membrane?
A) Group translocation
B) Active transport
C) Facilitated diffusion
D) Simple passive diffusion
Answer: D) Simple passive diffusion
Explanation: Small nonpolar molecules like O₂ and H₂O cross via passive diffusion.
72. Which component from Halobacterium salinarum absorbs sunlight and generates proton motive force (pmf) through photocycling under anaerobic or microaerobic conditions?
A) Chlorosomes
B) Bacteriochlorophyll a
C) Bacteriorhodopsin
D) Photosystem I
Answer: C) Bacteriorhodopsin
Explanation: Bacteriorhodopsin is a light-driven proton pump generating PMF without photosystems.
73. One of the signaling molecules used for bacterial cell-cell communication is:
A) N-acetyl hydroxy laccase
B) N-acyl homoserine laccase
C) N-acetyl homoserine lactone
D) N-acyl homoserine lactone
Answer: D) N-acyl homoserine lactone
Explanation: These molecules are used in quorum sensing, especially in Gram-negative bacteria.
74. The embryonic origin of squamous cell carcinoma is:
A) Endoderm
B) Mesoderm
C) Mesenchymal tissue
D) Ectoderm
Answer: D) Ectoderm
Explanation: Squamous epithelium arises from ectoderm, so do most carcinomas of skin and mucosa.
75. Kidney transplantation from one individual to another of the same species but genetically different is a:
A) Isograft
B) Allograft
C) Autograft
D) Xenograft
Answer: B) Allograft
Explanation: Allograft = between same species, different genetic makeup (e.g., organ donation).
76. When you are continuously exposed to a specific antigen for a longer period, which of the following statements is wrong?
A) There will be antibody class switching happening in the body
B) There will be affinity maturation happening in the body
C) The body produces the same antibody always
D) There will be primary and secondary response happening
Answer: C) The body produces the same antibody always
Explanation: Due to somatic hypermutation and class switching, the body produces different, higher-affinity antibodies over time.
77. Diffusion, formation of stable concentration gradients, and functioning of hedgehog proteins critically depend on their association with:
A) Fatty acids
B) Cholesterol
C) Linolenic acid
D) Carboxylic acid
Answer: B) Cholesterol
Explanation: Hedgehog proteins are post-translationally modified with cholesterol, crucial for proper signal gradient formation.
78. The visceral arches IV to VII of the basic vertebrate body plan gave rise to which of the following systems during evolution?
A) Digestive system
B) Respiratory system
C) Excretory system
D) Nervous system
Answer: B) Respiratory system
Explanation: These arches evolved into gill/respiratory structures, and in tetrapods, into parts of larynx and pharynx.
79. Widely used molecular characterization technique for identification of different strains of bacteria is:
A) Serological identification analysis
B) 5S ribosomal RNA gene analysis
C) 16S ribosomal RNA gene analysis
D) Total mRNA gene analysis
Answer: C) 16S ribosomal RNA gene analysis
Explanation: 16S rRNA gene is highly conserved and used for bacterial taxonomy and phylogeny.
80. Translocation in Philadelphia chromosome:
A) 19 to 21 chromosome region exchange
B) 9 to 21 chromosome region exchange
C) Leads to the production of black urine
D) Leads to chronic myelogenous leukemia
Answer: D) Leads to chronic myelogenous leukemia
Explanation: The Philadelphia chromosome (t(9;22)) creates BCR-ABL fusion gene, associated with CML.
81. Paradox related to protein folding is:
A) C-Value paradox
B) Sherman paradox
C) Lombard’s paradox
D) Levinthal’s paradox
Answer: D) Levinthal’s paradox
Explanation: Levinthal's paradox states that proteins fold extremely fast despite a huge number of possible conformations.
82. Glucose-6-phosphate and glyceraldehyde-3-phosphate are processed by glycolysis. How many ATP molecules will be generated respectively from each?
A) 2 ATP and 3 ATP
B) 3 ATP and 3 ATP
C) 3 ATP and 2 ATP
D) 4 ATP and 2 ATP
Answer: A) 2 ATP and 3 ATP
Explanation:
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G6P enters glycolysis early → 2 net ATP
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G3P is downstream → generates 3 ATP (if not used elsewhere)
83. How many ATP molecules are formed when sucrose is metabolized during glycolysis by substrate-level phosphorylation?
A) 2
B) 4
C) 8
D) 0
Answer: B) 4
Explanation: Sucrose = glucose + fructose → each gives 2 ATP via glycolysis → total 4 ATP (substrate-level).
84. The vitamin which plays an important role in conversion of proline to hydroxyproline in collagen synthesis is:
A) Vitamin A
B) Vitamin B
C) Vitamin C
D) Vitamin D
Answer: C) Vitamin C
Explanation: Vitamin C (ascorbic acid) is a cofactor for prolyl hydroxylase, essential for collagen stability.
85. In eukaryotes, the promoter sequence TATA box is located at:
A) 10–15 base pairs upstream of transcription site
B) 10–15 base pairs downstream of transcription site
C) 25–35 base pairs upstream of transcription site
D) 25–35 base pairs downstream of transcription site
Answer: C
Explanation: The TATA box is found ~25–35 bp upstream of the transcription start site and helps RNA polymerase binding.
86. Which one is not true in case of CpG?
A) CpG are island promoters
B) CpG are involved in gene regulation at chromatin level
C) CpG within promoters are methylated
D) Methylation in CpG is associated with 'C'
Answer: C
Explanation: In active promoters, CpG islands are typically unmethylated. Methylation silences gene expression.
87. Ubiquitin ligases are associated with degradation of:
A) Protein
B) DNA
C) RNA
D) Carbohydrates
Answer: A) Protein
Explanation: Ubiquitin ligases (E3) tag proteins for degradation via the proteasome pathway.
88. Which one of the following is not readily absorbed in the small intestine?
A) Na⁺
B) Cl⁻
C) K⁺
D) Mg²⁺
Answer: D) Mg²⁺
Explanation: Magnesium is absorbed less efficiently than Na⁺, Cl⁻, or K⁺ in the small intestine.
89. The menstrual cycle of human female is an example of biorhythm known as:
A) Ultradian rhythm
B) Tidal rhythm
C) Infradian rhythm
D) Circadian rhythm
Answer: C) Infradian rhythm
Explanation: Infradian rhythms last more than 24 hours (e.g., menstrual cycle ~28 days).
90. Which of the following heat shock proteins is present in almost every compartment of the eukaryotic cell?
A) Hsp60
B) Hsp70
C) Hsp90
D) Hsp100
Answer: B) Hsp70
Explanation: Hsp70 is ubiquitous, found in cytoplasm, nucleus, ER, and mitochondria, helping in protein folding.
91. Muscle overactivity like long running needs more O₂ to muscles. This is achieved by release of more O₂ from blood hemoglobin due to:
A) More blood flows to the muscles
B) Accumulation of lactic acid and increasing pH
C) Accumulation of lactic acid and reducing pH
D) Increased demand of O₂ by muscles
Answer: C
Explanation: Lactic acid reduces pH, shifting oxyhemoglobin dissociation curve → more O₂ released (Bohr effect).
92. Given the abbreviations, which equation best describes glomerular filtration rate (GFR)?
Kf – glomerular capillary filtration coefficient
Pg – glomerular hydrostatic pressure
Pb – hydrostatic pressure in Bowman’s capsule
Gc – osmotic pressure of plasma proteins
Bc – osmotic pressure in Bowman’s capsule
A) GFR = Kf × (Pg − Pb − Gc + Bc)
B) GFR = Kf + (Pg + Pb + Gc + Bc)
C) GFR = Kf − (Pg − Pb + Gc − Bc)
D) GFR = Kf × (Pg + Pb − Gc − Bc)
Answer: A
Explanation:
GFR = Kf × (Pg − Pb − πGc)
Here Bc is usually negligible or zero.
93. Name the water-soluble, non-B complex vitamin that is structurally similar to a monosaccharide:
A) Pantothenic acid
B) Biotin
C) Niacin
D) Ascorbic acid
Answer: D) Ascorbic acid
Explanation: Vitamin C (ascorbic acid) is structurally a lactone of glucose → water-soluble, not B-complex.
94. The amino acid involved as a precursor in auxin biosynthesis is:
A) Serine
B) Tryptophan
C) Valine
D) Tyrosine
Answer: B) Tryptophan
Explanation: Indole-3-acetic acid (IAA), a major auxin, is synthesized from tryptophan.
95. In aerobic respiration, oxidation of 1 molecule of glucose gives rise to how many ATPs (approximate)?
A) 10
B) 25
C) 30
D) 36
Answer: D) 36
Explanation: Approximate ATP yield per glucose via glycolysis, Krebs cycle, and oxidative phosphorylation is ~36.
96. For synthesis of 1 glucose molecule via the Calvin cycle, the plant requires:
A) 6 CO₂ + 30 ATP + 12 NADPH
B) 6 CO₂ + 12 ATP
C) 6 CO₂ + 18 ATP + 12 NADPH
D) 6 CO₂ + 18 ATP + 16 NADPH
Answer: C
Explanation: Calvin cycle needs 18 ATP and 12 NADPH to fix 6 CO₂ into 1 glucose molecule.
97. The folding of sheet of cells, cell migration, and apoptosis are mechanisms of:
A) Cleavage pattern
B) Pattern formation
C) Morphogenesis
D) Differentiation
Answer: C) Morphogenesis
Explanation: Morphogenesis defines shape and structure of tissues/organs via cell movement, death, and shape change.
98. In Drosophila, the fate of early blastomeres is determined through:
A) Autonomous specification
B) Conditional specification
C) Syncytial specification
D) Maternal specification
Answer: C) Syncytial specification
Explanation: In early Drosophila, nuclei share cytoplasm; morphogens like Bicoid influence fate before cellularization.
99. How does p53 prevent unrestrained proliferation of cells leading to cancer?
A) By preventing proliferation of cells with damaged DNA
B) By acting as a transcription factor
C) By stimulating synthesis of DNA repair enzymes which replace telomeres
D) By preventing cells from triggering apoptosis
Answer: A
Explanation: p53 checks DNA damage, arrests cell cycle or induces apoptosis to avoid propagation of mutations.
100. Match the correct combination between A and B:
| A | B |
|---|---|
| Prokaryotic cells | (iv) Less than 5 µm, lack membrane around all organelles |
| Eukaryotic cells | (i) More than 10 µm, no membrane around all organelles |
| Eukaryotic cells | (ii) More than 10 µm, lack membrane around some organelles |
| Prokaryotic cells | (iii) Less than 10 µm, possess membrane around some organelles |
A) 1 and (ii)
B) 2 and (i)
C) 3 and (iii)
D) 4 and (iv)
Answer: D
Explanation: Prokaryotes are typically <5 µm, and lack membrane-bound organelles, matching option D.
Why Solve MH SET 2023 Life Science Question Paper?
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Understand the Exam Pattern
Solving the MH SET 2023 Life Science paper gives you a clear idea of the type of questions asked, their difficulty level, and how topics are distributed across the syllabus. -
Identify Repeated Topics
Many core concepts in Life Science—like molecular biology, ecology, cell structure, biotechnology, and physiology—appear regularly in the exam. Practicing past papers helps you spot patterns and recurring topics. -
Improve Time Management
The SET exam is time-bound. Solving actual questions from 2023 lets you practice under real conditions, helping you improve speed and accuracy. -
Build Confidence
The more familiar you are with the question format and subject trends, the more confident and less anxious you'll feel on exam day.
How to Use These Solved Questions Effectively
This post is more than just a list of questions and answers—it’s a complete practice guide for MH SET aspirants. Here’s how to make the most of it:
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Attempt First, Then Check: Try solving each question on your own before looking at the explanation. This helps assess your preparation level.
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Read Explanations Thoroughly: Don’t skip the reasoning behind the answers. Each explanation is designed to reinforce your conceptual understanding.
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Make Notes of Important Points: Highlight keywords, formulas, and tricky concepts. These notes will be valuable during your final revision.
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Revise Frequently: Revisit the questions you struggled with. Focus on understanding the logic so you don't repeat the same mistakes.
